Without further discussion!!!Depending on the KOH concentration and temperature these are conditions that will produce benzyne, water may re-add but benzyne is so reactive that it will poylmerise rapidly. So at best you will get a mess.

Certainly, you could write that as your solution. If you could verify this as well, it would be even more convincing. I have neither. Being unaware of the benzyne polymerization rate and if I were attempting to perform this reaction and I was getting polymerization, I would doubtlessly slowly add the iodobenzene to the KOH to reduce the benzyne concentrations and increase the concentration of hydroxide. That should increase the bimolecular reaction rate.

This looks like a reaction scheme taken from an example of a benzyne reaction in which the objective is to predict the benzyne products. I presumed that was the point of using a dideuterated starting material.

Benzyne is so reactive that no matter how low the concentration was (is) you would get a mess, unless you used Schlenk conditions and trapped it out with a dienophile at low temp.

It has been shown (A. T. Bottini and J. D. Roberts, J. Am. Chem. Soc., 79, 1458 (1957).) that two mechanisms, elimination-addition (benzyne) and SN2 displacement, are operative in the liquid-phase hydrolysis of halogenated aromatic compounds. The formation of isomeric phenols as a result of the availability of the benzyne route makes the reaction of limited synthetic value. The incorporation of the copper-cuprous oxide system suppresses reaction via the benzyne route, so that the present method has general utility for the preparation of isomer-free phenols. For example, p-cresol is the only cresol formed from p-bromotoluene under the conditions of this preparation.